myrmepropagandistFeb 20

@Bumblefish

Which one is random?
(data sets are 100 numbers 1 to 6)

listA=[2,3,5,1,2,2,4,2,4,5,2,3,3,4,5,6,4,2,6,2,2,1,3,4,5,5,6,3,3,6,1,4,2,1,4,5,2,2,3,3,3,5,6,3,2,4,5,5,1,1,1,6,1,4,3,5,5,3,1,1,1,6,1,4,6,6,3,6,6,2,4,4,4,5,1,5,6,2,6,1,1,2,4,2,2,3,4,4,5,6,1,3,3,3,5,4,6,5,1,6]

listB=[4,2,5,6,3,5,3,1,3,4,2,3,4,3,4,5,5,1,3,3,2,1,1,6,1,3,2,2,2,6,1,5,6,3,6,3,2,3,2,4,6,1,1,6,3,2,4,1,6,1,3,1,5,6,2,3,3,5,1,6,4,5,2,5,1,1,5,3,6,2,3,3,6,5,2,3,3,1,6,3,2,3,2,1,6,6,4,4,6,2,4,5,4,5,3,4,6,5,3,2]

Show threadDaniel Lakeland

@futurebird
just to clarify what she means is as if from random unbiased 6 sided die rolls.

@Bumblefish

Feb 20 at 11:55pmWeb
Show threadDaniel LakelandFeb 21

@futurebird
things I would check are first the frequency of each number... they should be somewhat uniform but not TOO close to equal as all exactly equal is unlikely... next I'd look at the length of repeat sequences and compare to expected values.

the actual definition of random sequences (Per Martin-Löf) is in terms of passing tests actually
@Bumblefish

Show threadDianeFeb 21

@dlakelan @futurebird

The dictionaries in the Counter() object are the number of times each integer appears.

In [18]: Counter(listA)
Out[18]: Counter(
{2: 17, 3: 17, 5: 16, 1: 17, 4: 17, 6: 16}
)

In [19]: Counter(listB)
Out[19]: Counter(
{4: 12, 2: 17, 5: 14, 6: 17, 3: 24, 1: 16}
)

Show threadDaniel LakelandFeb 21

@alienghic
I'm on my phone at a volleyball game but what's the likelihood for each (probability of seeing that vector of counts given a multinomial distribution with 1/6 as probability for each value)

should be pretty easy in R or Julia or Python though offhand I would need to look at docs for any of them. Julia would be something like
using Distributions
pdf(Multinomial([1/6, 1/6,...], [17,17,17,17,16,16])
@futurebird

Show threadPensionDanFeb 21
@dlakelan @futurebird @Bumblefish Based on this description, A looks too uniform. B could be random.
Show threadDaniel LakelandFeb 21
@danpmoore
agreed, the frequencies seem too uniform for the first intuitively.
@futurebird @Bumblefish
Show threadVicente ⁂Feb 21

@dlakelan @futurebird @Bumblefish another thing to look for could be frequency of pairs of numbers. for an unbiased, independent dice, there should be about a 1/36 chance of each pair of numbers to appear.

unfortunately you'd quite a large number of randomly generated samples to get this chance exactly, but i guess you could do some fancy statistics to analyze these distributions and try to guess which one is "more random looking"